# Finding the WORK done by a crane in lifting a load

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1. Mar 18, 2016

### vinci

1. The problem statement, all variables and given/known data
A crane lifts its 500N load to the top of the building from A to B. Distances are shown on the diagram. Calculate how much work is done by the crane

Known variables:
AB= displacement= 50m
AC=horizontal distance=30m
BC=vertical distance=40m

2. Relevant equations

Work done= force * distance covered in direction of force
W=F.S

Force=mass* acceleration
F=ma

Hypotenuse= Root ( Adjacent^2 + Opposite^2 )

3. The attempt at a solution

From the diagram it can be seen that there the DISPLACEMENT is AB or 50 M. The load is 500N which will require a force in upward direction of 500N. I am confused with which variables to use but nevertheless
Work done in upward direction(BC):
W=FS
=500N*40M=20000NM

Work done in horizontal direction(AC):
W=FS
=500N*30M=15000NM I'm unsure about this one if 500N force will be required because we are not pushing a body on the ground and not countering a friction of 500N

Work done in AB?
W=FS
=500N * 50M =25000NM
(This can't be right can it? Because from what I think the crane will first lift the weight up and then rotate)
This is a simple highschool question and I might be overcomplicating it but I am really confused which variables to use since the question should have ONE answer.

2. Mar 18, 2016

### BvU

Hello Vinci,

Look carefully at your first relevant equation ! The only force the crane exerts on the load is in which direction ?

3. Mar 18, 2016

### vinci

The crane exerts a force in two directions. An upward one and then a horizontal one. Am I right?

4. Mar 18, 2016

### BvU

You can ignore the horizontal one: the sideways movement can be executed with negligibly small force (in theory. By doing it real slow and having a minimum of friction - something that isn't extremely realistic, but that's what they mean here).

5. Mar 18, 2016

### vinci

So I should only be concerned with the motion on y axis or the one due to 500N force and the distance covered by it is 40M making the answer 20000NM, right?
Why has the questioniare even mentioned the two other distances then/

6. Mar 18, 2016

### BvU

To make you think thoroughly about this "in the direction of the force" principle .

7. Mar 19, 2016

### Ahmer

I solved this question from the AS level book by calculating angle (theta) as in by tan of theta n then apply the Fdcos@ equation taking d as 50
Coz tension is working against gravity no? So the crane is moving both vertically and horizontally

8. Mar 19, 2016

### vinci

@Ahmer: cosTHEETA calculates the horizontal component of the force. I don't believe that's the right answer.

9. Mar 19, 2016

### BvU

Tension is pulling against gravity. Physically, the horizontal component is not doing any work because the displacement comes for free: there is no opposing force (no horizontal component of gravity -- and we agree to ignore the horizontal acceleration).